### Video instructions and help with filling out and completing How 8850 Form Hires

### Instructions and Help about How 8850 Form Hires

Music hi I'm Edgar and I'm a software engineer at Google hi I'm Becky and I'm a software engineer at Google so Edgar the question I'm going to give you today is a I'm going to give you a collection of numbers and I need you to take this collection of numbers and find a matching pair that is equal to a sum that I give you as well okay so for example the collection of numbers could be 1 2 3 & 9 and the sum that I'm looking for is 8 okay and then another example just for another set of numbers could be a 1 a 2 a 4 and a 4 and then again is some that I'm looking for is 8 so in this case there I guess what I'm trying to figure out is you're looking for a pair of numbers then that add up to 8 yeah right so in this case there isn't a pair of numbers that add up to it that is true ok example and in this case it is because the foreign for add up to 8 correct okay so this is this would be like no this is yes ok yes you ultimately happen okay so how are these numbers given can I assume that they're kind like in memory an array or something yeah they're in memory you can go with an array you can also assume that they're ordered intending water ok Oh interesting okay so how about repeating elements can I assume that they would be like for instance here what if I I guess but what if I didn't have that for what I use like the 4 and the 4 to get that 8 you can't repeat the same element at the same index twice but certainly the same number may appear twice ok ok so like that would be would be I yes how about these numbers are they integers or are they floating point or you can assume they'll always be users ok negatives positives negatives can happen ok cool so well the first the simplest solution of course is just comparing every single possible pair so I could just have two for loops one scanning the whole thing and then the second one starting from let's say you have the I loop and then the J loop starting from I plus one so that I don't repeat the same value and just testing all of them if the sum is equal to the to the target sum I mean that's obviously not very efficient but that would be like a way to solve it that would work it certainly would be time-consuming I think yes so any kind of example that yeah yeah I think that that would be quadratic so better than quadratic well since it's sort it's okay I guess I need to figure out when I have a number what I'm looking for is if there's another number that sounds too 8 so so if I have a 1 what I'd need to figure out is if there's a 7 somewhere in the array and that's the case it's sorted then I can just do binary search I guess if I go here and I binary search for a 7 then I go here and I binary search for a 6 which is the complement of that and when I go here I binary search for a 5 and at the end I just don't do anything and so in this case I would solve it like that so that's a bit better than quadratic I guess binary search is log algorithm in a sorted list also any answer ok you're kind of slow okay so what if you took a look at instead of doing a binary search which is unidirectional what if you started with a pair of members to begin with okay and then work your way through in work from there let's see so it's right okay let me try to bound this thing so the the largest possible some I guess would be the last two values that would be a largest possible Sun yeah the smallest possible Sun would be the two smallest right so so anything in between Wow okay so the range of the possible values is that right so there's nothing that is probably small there's nothing that can be smaller than this value right there's nothing that can be larger than that value that's okay so if this sum is 10 in this case it's too large so I need to find a smaller sum so I could just move this one over here and if that is too small now and I need to move that one over there okay so I can I think I can just do it with with that in a in a linear solution just moving at each iteration I either move the high one lower if I am if my pair is too large and I move my lower higher if my pair is too small and I end whenever I either find two like in this case I need to find a pair that adds up to eight or whenever they cross so everybody every point I'm moving one of them so they would have to at least cross and I move exactly one so that means that it's linear yeah so that that would be a way of solving that problem and how does that how does it make that faster than a binary search okay so in the binary search case I was doing log or finding but I had to repeat that for every element that I was an N log and solution in this case I just need to do that moving scanning the one time so it's a linear solution so that's that's faster right right okay so before maybe you could.